Previous Exams che31301 mass transfer spring 2011 quiz #1 I
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Figure III-1. Concentration of solute as a function of time in compartments A and B. IV. Pulverized coal, which may be approximated as carbon spheres of radius R = 1 mm, is burned in a pure oxygen atmosphere at conditions such that carbon monoxide is produced. Oxygen is transferred to the particle surface by diffusion, where it is consumed in the reaction C + O2 CO Let N”A be the molar flux of O2 and N”B be the molar flux of CO, we have a) N”A = 2N”B b) N”A = 2N”B c) N”A = 0.5N”B(ANS) d) N”A = N”B IV. The “drug patch” shown in the figure below releases a water-soluble epidermal growth factor (species A) to repair a specific region of wounded tissue on the human body. A slow release of the drug is critical for regulating the rate of tissue repair. The drug layer (pure solute A) rests on top of a diffusion barrier. The diffusion barrier is essentially a micro-porous polymer material consisting of tiny parallel pores filled with liquid water (species B). The diffusion barrier controls the rate of drug release. The thickness (L), pore size (dpore), and porosity of the diffusion layer determine the dosage rate of the drug to the tissue directly beneath it. The maximum solubility of the drug in water is 1 mole/m3 at 25oC. The total surface area of the patch is 4 cm2, but the cross-sectional area of the pores constitute only 35% of the surface area for flux. The effective diffusion coefficient of the drug in the diffusion barrier is 1.810-11 m2/s. 
(9) If the thickness of the diffusion barrier (L) is 5 mm, determine the dosage rate in mole per second, assuming that the drug is instantaneously consumed once it exits the diffusion barrier and enter the body tissue. ____________
D = Data: Viscosity of water at 25oC is 0.9078 cP, viscosity of water at 37oC is 0.7074 cP. D = 1.86910-10 m2/s CHE31301 MASS TRANSFER Spring 2011 QUIZ #4
Note: Your answers must be correct to 3 significant figures and have the appropriate units. I. Consider the transport of glucose from capillary blood to exercising muscle tissue. As a basis consider 1 gram of tissue. The glucose consumption of the tissue is 0.015 mol/gs. Blood flow to the region is 0.01 cm3/gs. The arterial glucose concentration is 5 mol/cm3. The value of PmS based on capillary recruitment during exercise is 0.004 cm3/ gs. Using the CSTR model calculate: 1) the glucose concentration in the tissue space 1.5 mol/cm3 2) the glucose concentration in the exit blood. 3.5 mol/cm3 II. In a hot combustion chamber, oxygen diffuses through a stagnant film of air with thickness L to the carbon surface where it reacts to make CO and CO2. The mole fraction of oxygen at z = L is 0.21. The reaction may be assumed to be instantaneous. No reaction occurs in the gas film. The chamber is at 1.5 atm, 1000oK, and L = 0.1 m. The diffusivity of oxygen at these conditions is 0.35 cm2/sec. The following reaction occurs at the carbon surface (Gas constant = 0.08205 atmm3/kmoloK): 4C + 3O2 2CO + 2CO2 (3) a) NO2 =  NCO2 b) NO2 = 1.5 NCO2 c) NO2 = NCO2 d) None of the above
III. A mixture contains 35 mole % isobutane and 65 mole % isopentane is at 30 psia. The K values for these compounds can be obtained from ln K = A/T2 + B + C ln P where T is in oR and P is in psia Compound A B C Isobutane -1,166,846 7.72668 -.92213 Isopentane -1,481,583 7.58071 -.93159
CHE31301 MASS TRANSFER Spring 2011 QUIZ #5
The concentration of NO is reduced by passing the exhaust gases over the surface, and the rate of reduction at the catalyst is governed by a first order reaction of the form r"A(kmol/m2∙s) = k”1CA where k”1 = 0.08 m/s and A denotes NO 
As a first order approximation it may be assume that NO reaches the surface by one-dimensional diffusion through a thin gas film of thickness L that adjoins the surface. Consider a situation for which the exhaust gas is at 750oK and 1.2 bars and the mole fraction of NO is yA0 = 0.25 if DA,mix = 10-4 m2/s and the film thickness is L = 1 mm. (1) The molar flux of A (NO) can be obtained by integrating the following expression (A) NA =   (Ans.) (B) NA = 
(C) NA = cDAB (D) None of the above
(2) If the molar of A is given by NA = cDAB(yA0 yAL)/L determine yAL 0.139 II) A distillation column with 100 kmol/h feed of 60% A and 40% B produces a distillate product with xD = 0.95 and a bottom stream with xbot = 0.04 of the more volatile species A. CMO is valid and the equilibrium data is given by y = 
3) For total reflux, determine (numerically) the composition (y) of the vapor stream entering the second equilibrium plate from the top. 0.7525
4) For a reflux ratio of 2, and q = 0.5, determine the liquid composition (x) of the feed point (the intersection of the q-line and the operating lines). x = 0.53 III. A mixture contains 35 mole % isobutane and 65 mole % isopentane is at 30 psia. The K values for these compounds can be obtained from ln K = A/T2 + B + C ln P where T is in oR and P is in psia Compound A B C Isobutane -1,166,846 7.72668 -.92213 Isopentane -1,481,583 7.58071 -.93159
then the mole fraction of isobutane (iC4) in the liquid phase is 0.2402 IV) Consider a distillation column with the McCabe-Thiele diagram given in Figure Q4-III. The column has a total condenser and a partial reboiler. Note: the points A, C, E, a, c, and e are on the equilibrium curve and the points B, D, F, b, d, and f are on the operating lines. 
Figure Q4-III McCabe-Thiele diagram for binary distillation. 6) A) The x coordinate of point D gives the mole fraction of the volatile species in the liquid stream entering equilibrium tray 4 from the top (with the top tray as tray 1). B) The y coordinate of point C gives the mole fraction of the volatile species in the vapor stream entering equilibrium tray 3 from the top. a. A and B are true. b. Only A is true (A) c. Only B is true d. A and B are false 7) A) The y coordinate of point d gives the mole fraction of the volatile species in the vapor stream entering equilibrium tray N-2 (with the bottom tray before the reboiler as tray N). B) The x coordinate of point d gives the mole fraction of the volatile species in the liquid stream leaving tray N-2. a. A and B are true (A). b. Only A is true c. Only B is true d. A and B are false 8) A) The feed is introduced into tray 6 from the top. B) The feed is introduced at the optimum location in the column. a. A and B are true. b. Only A is true c. Only B is true (A) d. A and B are false V) A liquid feed at the boiling point contains 5 mol % of species A and 95 mol % water and enters the top tray of a stripping tower shown in Figure Q4-V. Saturated steam is injected directly into liquid in the bottom of the tower. The overhead vapor which is withdrawn contains 98% of A in the feed. Assume equimolar overflow for this problem. Equilibrium data for mole fraction of A is given by y = 10x for x 0.08. 
(9) For an infinite number of theoretical steps, calculate the minimum moles of steam needed per mole of feed. Figure Q4-V Stripping tower Minimum 0.098 mol steam/mol feed (10) Using 15 moles of steam per 100 moles of feed, calculate the mole fraction of A in the vapor 0.3267 Download 125.86 Kb. Share with your friends:
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